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    bart10127's Avatar
    bart10127 Posts: 1, Reputation: 1
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    #1

    Oct 6, 2009, 02:20 PM
    Probability homework
    If you are randomly sampling one at a time with replacement from a bag of marbles that contains 8 blue marbles, 7 red marbles and 5 green marbles, what is the probability of obtaining at least 2 red marbles in 3 draws from the bag. I know if I needed the first 2 to be red (with replacement) I would use:

    (7/20)(7/20)= 49/400 = .1225

    I am confused with this 3 draws piece. Am I using p(a)p(b)p(c)?

    Any help showing me the right direction would be appreciated.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Oct 6, 2009, 11:06 PM

    Picking at least two red balls from the 3 first draws means that you have many different possibilities, namely,

    RRR
    RBR
    RGR
    GRR
    BRR
    RRG
    RRB

    as combinations. (R-red marble, B-blue marble, G-green marble)

    Find each probability, and add them up, because it is then and 'either-or' case.

    I hope this helped! :)
    liling's Avatar
    liling Posts: 2, Reputation: 1
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    #3

    Oct 8, 2009, 08:09 AM
    Answer question10-11 based on the following: A jar contains 15 lollipopd: 6 red, 2 green, 4 blue, and 3 yellow.

    1. A blue followed by a red. (2 lollipops are chosen with replacement.)
    liling's Avatar
    liling Posts: 2, Reputation: 1
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    #4

    Oct 8, 2009, 08:16 AM

    11. At least one green. (three lollipops are chosen with replacement.)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Oct 8, 2009, 08:19 AM

    1. Multiply the probability of having a blue lollipop by that of having a red lollipop.

    11. Take the probability of having no green at all from the three you take. Then, subtract that from 1.
    Chris-infj's Avatar
    Chris-infj Posts: 31, Reputation: 4
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    #6

    Oct 9, 2009, 11:29 AM
    Quote Originally Posted by bart10127 View Post
    If you are randomly sampling one at a time with replacement from a bag of marbles that contains 8 blue marbles, 7 red marbles and 5 green marbles, what is the probability of obtaining at least 2 red marbles in 3 draws from the bag. I know if I needed the first 2 to be red (with replacement) I would use:

    (7/20)(7/20)= 49/400 = .1225

    I am confused with this 3 draws piece. Am I using p(a)p(b)p(c)?

    Any help showing me the right direction would be appreciated.
    If you are sampling WITH replacement, then this means that the probability of drawing red in EACH draw is a constant (7/20) and of not drawing red is also a constant (13/20)

    Then if X is defined as the number of red marbles in three draws,

    X follows a binomial distribution with p = 7/20 and q = 13/20 and n = 3

    The general formula is P(X=r) = nCr x q^(n-r) x p^r

    For P(X is at least 2), you will need P(X=2) + P(X=3)

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